Find the heights to which water will in a capillary tube of radius 0.3mm, `sigma = 7.2 xx 10^(-2) Nm^(-1)` . What will happen if the tube is raised till the upper end is 3cm above the water surface?

Explanation

`h = ( 2 sigma cos theta )/( r rho g) = ( 2xx 7.2 xx 10^(-2) xx cos theta )/( 3 xx 10^(-4) xx 10^(3) xx 9.8) = 0.0 489 m`
In the second case water will rise upto 3cm with a new value for the radius of the meniscus. Let the new height be h. and the radius of curvature R. . Then
`R.h. = rh, R. = ( rh )/( h.) = ( 0.3 xx 10^(-3) xx 0.0 489 )/( 3 xx 10^(-2)) = 0.48 xx 10^(-3) m`